Description
Given four lists A, B, C, D of integer values, compute how many tuples (i, j, k, l) there are such that A[i] + B[j] + C[k] + D[l] is zero.
To make problem a bit easier, all A, B, C, D have same length of N where 0 ≤ N ≤ 500. All integers are in the range of -228 to 228 - 1 and the result is guaranteed to be at most 231 - 1.
Example:
Input:
A = [ 1, 2]
B = [-2,-1]
C = [-1, 2]
D = [ 0, 2]
Output:
2
Explanation:
The two tuples are:
1. (0, 0, 0, 1) -> A[0] + B[0] + C[0] + D[1] = 1 + (-2) + (-1) + 2 = 0
2. (1, 1, 0, 0) -> A[1] + B[1] + C[0] + D[0] = 2 + (-1) + (-1) + 0 = 0
这道题是之前那道 4Sum 的延伸，让我们在四个数组中各取一个数字，使其和为0。那么坠傻的方法就是遍历所有的情况，时间复杂度为 O(n4)。但是既然 Two Sum 那道都能将时间复杂度缩小一倍，那么这道题使用 HashMap 是否也能将时间复杂度降到 O(n2) 呢？答案是肯定的，如果把A和B的两两之和都求出来，在 HashMap 中建立两数之和跟其出现次数之间的映射，那么再遍历C和D中任意两个数之和，只要看哈希表存不存在这两数之和的相反数就行了，参见代码如下：
解法一：
class Solution {
public:
int fourSumCount(vector<int>& A, vector<int>& B, vector<int>& C, vector<int>& D) {
int res = 0;
unordered_map<int, int> m;
for (int i = 0; i < A.size(); ++i) {
for (int j = 0; j < B.size(); ++j) {
++m[A[i] + B[j]];
}
}
for (int i = 0; i < C.size(); ++i) {
for (int j = 0; j < D.size(); ++j) {
int target = -1 * (C[i] + D[j]);
res += m[target];
}
}
return res;
}
};
下面这种方法用了两个 HashMap 分别记录 AB 和 CB 的两两之和出现次数，然后遍历其中一个 HashMap，并在另一个 HashMap 中找和的相反数出现的次数，参见代码如下：
解法二：
class Solution {
public:
int fourSumCount(vector<int>& A, vector<int>& B, vector<int>& C, vector<int>& D) {
int res = 0, n = A.size();
unordered_map<int, int> m1, m2;
for (int i = 0; i < n; ++i) {
for (int j = 0; j < n; ++j) {
++m1[A[i] + B[j]];
++m2[C[i] + D[j]];
}
}
for (auto a : m1) res += a.second * m2[-a.first];
return res;
}
};
类似题目：
参考资料：
https://leetcode.com/problems/4sum-ii/
https://leetcode.com/problems/4sum-ii/discuss/93920/Clean-java-solution-O(n2)
https://leetcode.com/problems/4sum-ii/discuss/93925/Concise-C%2B%2B-11-code-beat-99.5